Showing a group contains an element of order $n$ if $G/N$ contains an element of order $n$.

Question

(Problem) If $N$ is finite normal subgroup of $G$ and $G/N$ contains an element of order $n$, then $G$ contains an element of order $n$.

Let $gN\in G/N$ be an element of order $n$. Then $(gN)^n=N$. Since $(gN)^n=g^nN$, $g^n\in N$. Since $N$ is finite, there is a smallest $m$ for which $(g^n)^m=1$. So $(g^m)^n=1$. I attempt to show $g^m$ is the element of $G$ of order $n$. Suppose there is $k<n$ with $g^{mk}=1$. Although $mk<mn$, $mn$ may not be the order of $g$. So how can I obtain a contradiction?

Answer 1

Well you are going on the right track. Few final touches required:

Let say $g^k=e$, $g^n=a\in N$ for $1\leq k < nm$

Now, $(gN)^k=g^kN= N$. We know that this implies $n|k$. So we can write $k=nt$ .

Now going back to the first statement $(g)^k=e\implies (g)^{nt}=e\implies {(g^n)}^{t}=e \implies a^t=e$.

This shows that $m|t$ . But this is a contradiction since we get $nm|k$ but we assumed that $k<nm$.

Answer 2

I had trouble reading the accepted answer, so I'm posting a rephrasing of the argument.

Suppose $G/N$ contains an element $gN$ of order $n$. Then $g^n N = N$ implies that $g^n = a \in N$, so $\left( g^{|a|} \right)^n = 1$. We claim that the order of $g^{|a|}$ is $n$. Indeed, if $\left( g^{|a|}\right)^\ell = 1$ for some positive integer $\ell$, then $(gN)^{|a| \ell} = N$ implies that $|a| \ell = nq$ for some positive integer $q$, so $1 = \left(g^{|a|} \right)^\ell = g^{nq} = a^q$ and thereby $|a|$ divides $q$, thus $\ell$ is a multiple of $n$.